= Solution
The displayed orbit has the normalization \b[$\alpha=1$]. To verify it put $z=\tanh(\tau/\sqrt2)$ and $f=1-z^2=\operatorname{sech}^2(\tau/\sqrt2)$. Then $z'=f/\sqrt2$, $f'=-\sqrt2 fz$, so
$$
u'=3\sqrt2 fz=v,\qquad
v'=3f(1-3z^2).
$$
Also $u=-2+3z^2$, and $(u^2-1)=3(1-z^2)(1-3z^2)$, exactly the same expression as $v'$. Therefore the conservative equations hold. As $\tau\to\pm\infty$, $(u,v)\to(1,0)$; at $\tau=0$ the orbit reaches $(-2,0)$. Its conserved energy is $2/3$, so this is the required <homoclinic orbit>.
For a general positive $\alpha$ set $s=\sqrt\alpha$. Rescaling the verified solution gives
$$
\boxed{u=s\left[1-3\operatorname{sech}^2\left(\frac{\sqrt s\,\tau}{\sqrt2}\right)\right],\quad
v=3\sqrt2\,s^{3/2}\operatorname{sech}^2\left(\frac{\sqrt s\,\tau}{\sqrt2}\right)
\tanh\left(\frac{\sqrt s\,\tau}{\sqrt2}\right).}
$$
Direct differentiation gives $u'=v$ and $v'=u^2-\alpha$, and the endpoints are $(s,0)$. This restores the parameter dependence implicit in the normalized printed orbit.
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